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  • codeforces 1277 D xg

    题意:给n个01字符串,让你进行k下操作,使得t字符串翻转,使得n个字符串存在一个顺序,使得ai+1的首字母和ai相等。使得k最小,输出这k个数。(注意变化前和变化后的字符串不重复)

    思路:只有四种字符串,记录下每种字符串的数量。如果11,00,都不为0,且10,01为0,则无解。

      考虑sum1和sum2的差值小于1是有解的,否则考虑使得数量多的。

      若10多于01,考虑10的字符串翻转状态不存在在原字符串的数量够不够(sum10-sum01)/2。

      字符串标记map就行。

    #include <iostream>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <map>
    #include <iomanip>
    #include <algorithm>
    #include <queue>
    #include <stack>
    #include <set>
    #include <vector> 
    // #include <bits/stdc++.h>
    #define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    #define sp ' '
    #define endl '
    '
    #define inf  0x3f3f3f3f;
    #define FOR(i,a,b) for( int i = a;i <= b;++i)
    #define bug cout<<"--------------"<<endl
    #define P pair<int, int>
    #define fi first
    #define se second
    #define pb(x) push_back(x)
    #define ppb() pop_back()
    #define mp(a,b) make_pair(a,b)
    #define ms(v,x) memset(v,x,sizeof(v))
    #define rep(i,a,b) for(int i=a;i<=b;i++)
    #define repd(i,a,b) for(int i=a;i>=b;i--)
    #define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
    #define sca2(a,b) scanf("%d %d",&(a),&(b))
    #define sca(a) scanf("%d",&(a));
    #define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
    #define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
    #define scall(a) scanf("%lld",&(a));
    
    
    using namespace std;
    typedef long long ll;
    ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
    ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
    ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}
    
    const double Pi = acos(-1.0);
    const double epsilon = Pi/180.0;
    const int maxn = 2e5+10;
    map<string,int>vis;
    string a[maxn];
    int ans[maxn];
    int main()
    {
        // freopen("input.txt", "r", stdin);
        int _;
        scanf("%d",&_);
        while(_--)
        {
            int n;
            cin>>n;
            int tail = 0;
            int sum1 = 0,sum2 = 0,sum11=0,sum00=0;
            queue<int>que1;
            queue<int>que2;
            rep(i,1,n){
                string s;
                cin>>s;
                a[i] = s;
                vis[s] = 1;
                int len = s.length()-1;
                if(s[0]=='0'&&s[len]=='0') sum00++;
                else if(s[0]=='0'&&s[len]=='1') {
                    que2.push(i);
                    sum2++;
                }
                else if(s[0]=='1'&&s[len]=='1') sum11++;
                else if(s[0]=='1'&&s[len]=='0') {
                    que1.push(i);
                    sum1++;
                }
            }
            if(sum11&&sum00&&sum1==0&&sum2==0){
                cout<<-1<<endl;
                continue;
            }
            else {
                if(sum1 >= sum2){
                    int cha = sum1 - sum2;
                    if (cha <= 1)
                    {
                        cout<<0<<endl;
                        /* code */
                    }
                    else {
                        cha /= 2;
                        tail = 0;
                        while(que1.size()){
                            if(tail == cha) break;
                            int pos = que1.front();
                            que1.pop();
                            string tmp = a[pos];
                            //int k = tmp.length();
                            reverse(tmp.begin(),tmp.end());
                            if(vis[tmp]==1) continue;
                            ans[++tail] = pos;
                        }
                        if(tail == cha){
                            cout<<tail<<endl;
                            rep(i,1,tail){
                                cout<<ans[i]<<sp;
                            }
                            cout<<endl;
                        }
                        else cout<<-1<<endl;    
                    }
                }
                else {
                    int cha = sum2 - sum1;
                    if (cha <= 1)
                    {
                        cout<<0<<endl;
                        /* code */
                    }
                    else {
                        cha /= 2;
                        tail = 0;
                        while(que2.size()){
                            if(tail == cha) break;
                            int pos = que2.front();
                            que2.pop();
                            string tmp = a[pos];
                            reverse(tmp.begin(),tmp.end());
                            if(vis[tmp]==1) continue;
                            ans[++tail] = pos;
                        }
                        if(tail == cha){
                            cout<<tail<<endl;
                            rep(i,1,tail){
                                cout<<ans[i]<<sp;
                            }
                            cout<<endl;
                        }
                        else cout<<-1<<endl;
                        
                    }cout<<endl;
                }            
            }
            rep(i,1,n){
                vis[a[i]] = 0;
            }
            //cout<<sum1<<sp<<sum2<<endl;
    
        }
    }
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  • 原文地址:https://www.cnblogs.com/jrfr/p/13266736.html
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